diff --git a/EDIT2013_Kondo_ans/Makefile b/EDIT2013_Kondo_ans/Makefile
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+all:
+	pdflatex kondo_ex.tex
+
+clean: 
+	rm -rf *.aux kondo_ex.pdf *.log
diff --git a/EDIT2013_Kondo_ans/kondo_ex.tex b/EDIT2013_Kondo_ans/kondo_ex.tex
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+\documentclass[a4paper]{article}
+\usepackage[utf8x]{inputenc}
+\usepackage{ucs}
+\usepackage{amsmath}
+\usepackage{amsfonts}
+\usepackage{amssymb}
+\usepackage{epsf}
+\usepackage{float}
+\usepackage{url}
+\usepackage{fancyhdr}
+\usepackage{verbatim}
+\usepackage{color,listings}
+\usepackage{booktabs}
+\usepackage{tabularx}
+\usepackage{natbib}
+\usepackage[all]{xy}
+\usepackage{graphicx}
+\usepackage{graphics}
+\usepackage{color}
+\usepackage{listings}
+
+%\makeatletter
+%\def\@xobeysp{}
+%\makeatother
+\usepackage{color,listings}
+
+\author{Tran Hoai Nam}
+\title{Answer to quizzes in \\Detectors of LHC Experiments lecture}
+
+\begin{document}
+\maketitle
+\section*{Quiz 1: Number of protons  for $\mathcal{L}=10^{34}$
+cm$^{-2}$s$^{-1}$}
+Total luminosity of LHC:
+\[
+\mathcal{L} = f N_b L_{bunch\_cross} = f N_b \frac{N_p \times N_p}{4\pi\sigma_x\sigma_y}
+\]
+\[
+\rightarrow N_p = \sqrt{ 4\pi \sigma_x\sigma_y \frac{\mathcal{L}}{fN_b}}
+\]
+where: $f$ is revolution frequency; 
+\[
+f \approx \frac{c}{l_{LHC}} 
+= \frac{3\times 10^8 [ms^{-1}]}{27 \times 10^3 [m]} \approx 10^4 [s^{-1}]
+\]
+Substitute the numbers, number of proton in each bunch is:
+\[
+N_p = 
+\sqrt{4\times 3.14 \times 16 \times 10^{-6} [m]\times 16 \times 10^{-6} [m]
+\times
+\frac{10^{34} \times 10^{4}[m^{-2}s^{-1}]}{2808 \times 10^4[s^{-1}]}}
+\approx 1.07 \times 10^{-11} 
+\]
+
+\section*{Quiz 2: Characteristics of a silicon detector}
+\subsection*{1) Full depletion voltage}
+Let $V$ be the full depletion voltage for the 300 $\mu m$ Si detector:
+\[
+d = \sqrt{\frac{2\varepsilon^{'} \varepsilon_0}{q_e N_{eff}}V}
+\rightarrow  V = \frac{q_e N_{eff}}{2 \varepsilon^{'} \varepsilon_0} d^2
+\]
+\[
+V = \frac{1.6 \times 10^{-19} [C] \times 2 \times 10^{12} [cm^{-3}]}
+{2\times11.6\times 8.854 \times 10^{-14} [F cm^{-1}]} (300 \times 10^{-4} [cm])^2
+\approx 140.2 [V]
+\]
+\subsection*{2) Noise}
+Capacitance of the detector when it is fully depleted:
+\begin{align*}
+C &= \varepsilon^{'} \varepsilon_0 \frac{S}{d} \\
+ 	&= 11.6 \times 8.854 \times 10^{-14} [F cm^{-1}] 
+		\frac{1 [cm^2]}{300 \times 10^{-4} [cm]} \\
+  &\approx 34.2 \times 10^{-12} [F] \\
+	&= 34.2 [pF]
+\end{align*}
+
+Equivalent noise charge square:
+\[
+ENC^2 = \left(\frac{e^3}{36q_e}\right)^2 
+\left[ \frac{3}{\tau} (4k_B T R_S) C^2 + \frac{5\tau}{3}(2q_eI_{leak})\right]
+\]
+
+We have:
+\begin{align*}
+k_B T &= 8.62 \times 10^{-5} [eV K^{-1}] \times 273.15 [K] \\
+			&\approx  2.35 \times 10^{-2} [eV] \\
+			&= 2.35 \times 10^{-2} \times 1.6 \times 10^{-19} [J]\\
+			&= 2.35 \times 10^{-2} q_e [JC^{-1}]
+\end{align*}
+Thus:
+\begin{align*}
+ENC^2 = 
+	&	\frac{2.72^6}{36^2 \times 1.6 \times 10^{-19}[C]} \times \\
+	& \times \Biggl[ 
+		\frac{3}{20 \times 10^{-9} [s]} 
+		(4 \times 2.35 \times 10^{-2} [JC^{-1}] \times 200 [\Omega]) 
+		(34.2 \times 10^{-12} [F])^{2} + \\ 
+	& + \frac{5 \times 20 \times 10^{-9}[s]}{3}
+		(2 \times
+		 10 \times 10^{-6} [A])
+	\Biggr]
+\end{align*}
+
+\begin{align*}
+ENC^2 &= 
+	\frac{2.72^6}{36^2 \times 1.6 \times 10^{-19}[C]} 
+	 \times \Biggl[ 
+	3.3 \times 10^{-12}[Js^{-1}C^{-1}\Omega F^2] + 0.3 \times 10^{-12} [C]
+	\Biggr] \\
+	&\approx 7.0\times 10^6
+\end{align*}
+
+(dimensionally speaking, $Js^{-1}C^{-1}\Omega F^2
+= VA \times C^{-1}\times VA^{-1}\times C^2V^{-2} = C$)
+
+\[
+\rightarrow ENC \approx 2.65 \times 10^3
+\]
+
+The $ENC$ is $2.65 \times 10^3$, and it can be seen that the contribution from
+capacitance is dominant.
+%\begin{align*}
+%ENC^2 = 
+	%&\left(
+		%\frac{2.72^3}{36\times 1.6\times 10^{-19}[C]}
+	%\right)^2 \times \\
+	%& \times \Biggl[ 
+		%\frac{3}{20 \times 10^{-9} [s]} 
+		%(4 \times 8.62 \times 10^{-5} [eV K^{-1}] 
+		%\times 273.15 [K] \times 200 [\Omega]) 
+		%(34.2 \times 10^{-12} [F])^{2} + \\ 
+	%& + \frac{5 \times 20 \times 10^{-9}[s]}{3}
+		%(2 \times 1.6 \times 10^{-19}[C] \times
+     %10 \times 10^{-6} [A])
+	%\Biggr]
+%\end{align*}
+
+%\begin{align*}
+%ENC^2 = 
+	%&\left(
+		%\frac{2.72^3}{36\times 1.6\times 10^{-19}[C]}
+	%\right)^2 \times \\
+	%& \times \Biggl[ 
+		%\frac{3}{20 \times 10^{-9} [s]} 
+		%(4 \times 2.35 \times 10^{-2}[eV] \times 200 [\Omega]) 
+		%(34.2 \times 10^{-12} [F])^{2} + \\ 
+	%& + \frac{5 \times 20 \times 10^{-9}[s]}{3}
+		%(2 \times 1.6 \times 10^{-19} [C]\times
+     %10 \times 10^{-6} [A])
+	%\Biggr]
+%\end{align*}
+
+%\begin{align*}
+%ENC^2 = 
+	%&\left(
+		%\frac{2.72^3}{36\times 1.6\times 10^{-19}[C]}
+	%\right)^2 \times \\
+	%& \times \Biggl[ 
+		%\frac{3}{20 \times 10^{-9} [s]} 
+		%(4 \times 2.35 \times 10^{-2} \times 1.6 \times 10^{-19}[J] \times 200 [\Omega]) 
+		%(34.2 \times 10^{-12} [F])^{2} + \\ 
+	%& + \frac{5 \times 20 \times 10^{-9}[s]}{3}
+		%(2 \times 1.6 \times 10^{-19} [C]\times
+     %10 \times 10^{-6} [A])
+	%\Biggr]
+%\end{align*}
+
+\end{document}